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Alkanes, alkenes, and heat

  1. Jan 17, 2005 #1
    A hydrocarbon that has lots of double bonds takes more energy to melt, so therefore takes more heat than a hydrocarbon with less double bonds right?

    so a hydrocarbon with no double bonds will require the least amount of energy, so the last amount of heat

    is my concept right?

  2. jcsd
  3. Jan 17, 2005 #2


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    I'm sorry,but i don't see a direct connection between the number of double bonds in a carbon compound and the liquifying latent heat.The latter has to do with the crystaline structure of the solid hydrocarbon and the intermolecular bonds and not necessarily with the intramolecular bonds.

    Perhaps someone else could come up with the connection u're searching for...

  4. Jan 17, 2005 #3


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    Well, alkenes are generally more polarizable than alkanes, so you would expect a higher boiling point based on dispersion forces.
  5. Jan 17, 2005 #4
    Assuming that the hydrocarbon is in a liquid form then if you heat some of it it will produce smoke. If the smoke is 'smokey' then it has more alkenes in it. If the smoke is more whispy then it has more alkanes in it.

    This may not have been explained well.

    The Bob (2004 ©)
  6. Jan 17, 2005 #5
    ok, well here is the exact details of the problem:

    which fatty acid melts at the highest temperature (solid to liquid)

    (order from lowest to highest temp required to melt)
    Don't know how to draw on here, so i will just explain it to you

    1) a 17 long hydrocarbon chain with a Carboxyl group attached at the end. Has no double bond

    2) a 17 long hydrocarbon chain with a Carboxyl group attached at the end. Has 1 double bond

    3) a 17 long hydrocarbon chain with a Carboxyl group attached at the end. Has 2 double bonds.

    My guess is that 1 requires least and 3 requires most temperature
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